� � � � � � � � �� � � � � � � � � 18.03 Problem Set 7: Part II Solutions Part I points: 26. 6, 27. 10, 28. 12. I.26. e−t sin(3t) = 1 e(−1+3i)t (−1−3i)t , so L[e−t sin(3t)] = � 2i − e� 1 1 1=1 (s + 1 + 3i) − (s + 1 − 3i)= 3. 2i s − (−1 + 3i) − s − (−1 − 3i) 2i (s + 1)2 + 9 (s + 1)2 + 9 ∞ ∞ 26. (a) [10] G(s) = f(at)e−st dt. To make this look more like F (s) = f(t)e−st dt, 0 0 make the substitution u = at. Then du = a dt and ∞ f(u)e−su/a du 1 ∞ f(u)e−(s/a)u du =1 � s � G(s) = = F . 0 a a 0 a a n! ann! For example, take f(t) = tn, so F (s) = sn+1 , g(t) = (at)n = antn , G(s) = sn+1 . Now � � n+1 n1 s 1 n! a n! a n! compute F = = = = G(s). a a a (s/a)n+1 a sn+1 sn+1 (b) [10] Compute F (s)G(s) = ∞ ∞ f(x)e−sx g(y)e−sy dxdy = f(x)g(y)e−s(x+y) dxdy, 0 0 R where R is the first quadrant. The suggested substitution is x = t − τ, y = τ . To convert ∂x ∂x to these coordinates, note that the Jacobian is det ∂t ∂τ = det 1 −1 = 1 For ∂y ∂y 0 1∂t ∂τ fixed t, τ ranges over numbers between 0 and t, and t ranges over positive numbers. Since ∞ t x + y = t, F (s)G(s) = f(t − τ)g(τ)e−st dτdt � �� 0 �0 � �∞ t ∞ ∞ = f(t − τ)g(τ) dτ e−st dt = (f(t) ∗ g(t)) e−st dt = h(t)e−st dt = H(s). 0 0 0 0 � � 1 � ∞ ∞ (c) [6] F (s) = f(t)e−st dτ = f(t)e−st dτ + 0e−st dτ. The improper integral 0 0 1 converges for any s; the region of convergence is the whole complex plane. Continuing, �11 e−st �� 1 − e−s F (s) = −s � 0 = s . [Why do esn’t this blow up when s → 0? The numerator goes to zero too, then, and the limit of the quotient (by l’Hopital for example) is 1.] 27. (a) [4] The Laplace transform of the equation aw˙ +bw = δ(t) is asW (s)+bW (s) = 1. Solve: W (s) = 1= 1/a This is the Laplace transform of w(t) = a 1 u(t)e−bt/a. as + b s + (b/a) (b) This is called the “unit ramp response.” (i) [6] xp = c1t+c0, ac1+b(c1t+c0) = t, c1 = 1 b (as long as b = 0), � ac1+bc0 = 0 so c0 = −ba 2 , xp = 1 t− a x(t) = xp +ce−bt/a, so 0 = x(0) = −ba 2 +c and x(t) = u(t)(1 t− a (1−e−bt/a)).b b2 . b b2 If b = 0 then ax˙ = t, which has general solution x(t) = 21 a t2 + c. 0 = x(0) = c, so x(t) = u(t)21 a t2 . t t1 1 (ii) [6] If b �= 0: w(t) ∗ t = 0 ae−b(t−τ )/aτ dτ = ae−bt/a 0 ebτ/aτ dτ. Do this by parts:� � � � �� t � u = τ, du = dτ, dv = ebτ/adτ, v = ab ebτ/a, w(t) ∗ t = a 1 e−bt/a τ abebτ/a ��t 0 − 0 ab ebτ/a dτ = a 1 e−tb/a tab ebt/a ab22 (ebt/a − 1) = 1 b t − ba 2 (1 − e−bt/a).− t 1 1 t2 If b = 0, w(t) ∗ t = τ dτ = . 0 a a 2 1 A B C (iii) [6] ax˙+bx = t has Laplace transform asX+bX = 1 2 , so X = = + + ss2(as + b) s s2 as + b Coverup: Multiply by s2 and set s = 0 to get B = 1 . Multiply by as + b and set s = − b b a 2 to get C = ab2 . Here’s a clean way to get A: multiply through by s and then take s very a/b2 1/b a/b2 large in size. You find 0 = A + Ca , or A = −ba 2 . So X = −s + s2 + s + b/a, which is the Laplace transform of x = − a + 1 t + ba 2 e−bt/a.b2 b 1 1 1 If b = 0, ax˙ = t has Laplace transform asX = so X = , and x = u(t)1 1 t2 2 3 a 2s a s1 1 1 28. ( a) [6] w(t) has Laplace transform W (s) = 3s2 + 6s + 6 = 3 (s + 1)2 + 1 . L[sin t] = 1 , so by s-shift w(t) = 1 u(t)e−t sin t. s2+1 3 1 (b) [14] W (s) = . To use partial fractions we need to factor s4 + 1, which is to say s4 + 1 we need to find its roots. They are the fourth roots of −1, which are r, r, −r, and −r where 1 1 a b c d r = √12 (1+i). Now s4 + 1 =(s − r)(s − r)(s + r)(s + r)= s − r + s − r + s + r + s + r . Coverup or cross-multiplication will lead to the coefficients. This is not pretty, and (per the web) I don’t expect more. [What I intended to ask was for the weight function for D4 −I. Now the roots are ±1 and 1 a b c d ±i, so we can write = + + + . Coverup gives easily a = s4 s + 1 s + i 1 i ib = 4 , c = 4 , d = −4 . So− w1(t) s =−u1(t)14 (et + e−t s +−iei it − ie−it) = u(t)12 (sinh(t) − sin(t)). I apologize for the mistake.]MIT OpenCourseWare http://ocw.mit.edu 18.03 Differential Equations���� Spring 2010 For information about citing these materials or our Terms of Use, visit:
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